Continuous Aggregate Densities And Their Atom

Why an Aggregate Distribution Has Two Parts

An aggregate loss can be zero because no claims occur, while a positive aggregate is built by adding one or more claim amounts. When claim amounts have continuous distributions and are strictly positive, these possibilities have different mathematical forms: a probability mass at zero and a continuous density over positive loss amounts. Keeping the two parts separate is essential when calculating probabilities or describing the full distribution.

Let $N$ be the claim count, and let $X_1,X_2,\ldots$ be independent and identically distributed claim severities, independent of $N$. Assume each severity is strictly positive with probability one and has density $f_X$. Define the aggregate loss by $S=0$ when $N=0$, and $S=\sum_{j=1}^N X_j$ when $N\geq 1$. Write $p_n=P(N=n)$.

Definition: For $n\geq 1$, let $f_X^{*n}$ denote the density of the sum of $n$ independent severities, each with density $f_X$. In particular, $f_X^{*1}=f_X$, and $f_X^{*n}$ is the $n$-fold convolution of $f_X$. The distribution of $S$ has an atom of size $p_0=P(N=0)$ at zero and, for positive loss amounts, density $$ f_S(s)=\sum_{n=1}^{\infty}p_n f_X^{*n}(s), \qquad s>0. $$ If $N$ has a finite maximum value, terms beyond that value are zero.

This formula describes a mixture: conditional on $N=n$, the aggregate has density $f_X^{*n}$, and that density receives weight $P(N=n)$. The weights in the positive-loss density are the unconditional count probabilities $p_n$, not count probabilities rescaled to sum to one. The remaining weight $p_0$ belongs to the atom at zero.

The assumption that severities are strictly positive matters. It ensures $S=0$ exactly when $N=0$, so $P(S=0)=p_0$. If individual severities could themselves equal zero with positive probability, a positive claim count might still produce zero aggregate loss, and this simple identification would no longer hold.

Why the Mixture Formula Holds

Condition on the possible values of $N$. If $N=0$, then $S=0$. If $N=n\geq1$, then $S$ is a sum of $n$ independent severities and has density $f_X^{*n}$. For any set of positive loss amounts, its probability is therefore the sum of the conditional probabilities weighted by $P(N=n)$. This gives the mixture density.

Separate the zero-count case.
$P(N=0)=p_0$, and this entire probability is concentrated at $S=0$.

Condition on a positive count.
For each $n\geq1$, the sum of $n$ independent severities has density $f_X^{*n}$.

Weight and add the components.
The unconditional contribution from count $n$ is $p_n f_X^{*n}(s)$, so adding these contributions gives the density for $s>0$.

The same conditioning gives the cumulative distribution function. For $s\geq0$, let $F_X^{*n}(s)$ be the probability that a sum of $n$ severities is at most $s$. Then

P(S\leq s)=p_0+\sum_{n=1}^{\infty}p_n F_X^{*n}(s).

At $s=0$, every positive-count sum is strictly positive, so this expression gives $P(S\leq0)=p_0$. For $s<0$, the probability is zero. Integrating the positive-loss density over all $s>0$ gives $\sum_{n\geq1}p_n=1-p_0$; together with the atom, the total probability is one.

Worked Density Calculations

Worked Example: One Possible Claim

A hypothetical portfolio has claim count $N\sim\operatorname{Bernoulli}(0.4)$: it has one claim with probability $0.4$, and no claims otherwise. Given a claim, its severity is uniform on $(0,2)$. Find the atom at zero, the positive-loss density, and $P(S\leq1)$.

Method and assumptions. The severity is strictly positive and has density $f_X(x)=1/2$ for $0<x<2$, zero otherwise. The count probabilities are $p_0=0.6$ and $p_1=0.4$, with all other $p_n=0$.

The atom at zero is $P(S=0)=p_0=0.6$. Since only one claim can occur, the mixture has one positive-count term:

f_S(s)=p_1 f_X(s)= \begin{cases} 0.4(1/2)=0.2, & 0<s<2,\\ 0, & s>2. \end{cases}

The continuous component has total probability $\int_0^2 0.2\,ds=0.4$, which agrees with $P(N=1)$. To find $P(S\leq1)$, include the atom and the continuous probability from zero to one:

P(S\leq1)=0.6+\int_0^1 0.2\,ds=0.6+0.2=0.8.

Thus $P(S\leq1)=0.8000$, and the total probability is $0.6+0.4=1$. Common mistake avoided: using the continuous density alone would give $0.2$ for this probability and omit the $0.6$ chance of no claims.

Worked Example: Two-Claim Binomial Frequency

A hypothetical group has $N\sim\operatorname{Binomial}(2,0.5)$. Claim severities are independent and uniform on $(0,1)$, independent of $N$. Find the atom and density of $S$, and calculate $P(S\leq1)$.

Method and assumptions. The count probabilities are $p_0=0.25$, $p_1=0.50$, and $p_2=0.25$. A single severity has density one on $(0,1)$. For two severities, convolution gives

f_X^{*2}(s)= \begin{cases} s, & 0<s<1,\\ 2-s, & 1<s<2,\\ 0, & \text{otherwise}. \end{cases}

For example, when $0<s<1$, the convolution integral is $\int_0^s 1\cdot1\,dy=s$. For $1<s<2$, the overlap of the two unit intervals has length $2-s$. Weighting the one- and two-claim densities gives

f_S(s)= \begin{cases} 0.50+0.25s, & 0<s<1,\\ 0.25(2-s), & 1<s<2,\\ 0, & s>2. \end{cases}

The atom is $P(S=0)=p_0=0.25$. To check the continuous component, integrate each interval:

\int_0^1(0.50+0.25s)\,ds=0.50+0.125=0.625, \qquad \int_1^2 0.25(2-s)\,ds=0.125.

The positive density has total probability $0.625+0.125=0.75=p_1+p_2$, as required. For the requested probability,

P(S\leq1)=0.25+\int_0^1(0.50+0.25s)\,ds =0.25+0.625=0.875.

Hence $P(S\leq1)=0.8750$. Common mistake avoided: the two-claim density must be multiplied by $p_2=0.25$; adding the unweighted convolution would incorrectly treat two claims as certain.

Worked Example: A Density on a Restricted Interval

A hypothetical aggregate has $P(N=0)=0.1$, $P(N=1)=0.2$, $P(N=2)=0.3$, and $P(N=3)=0.4$. Severities are independent and uniform on $(0,1)$. Find the density of $S$ for $0<s<1$, then calculate $P(S<1)$.

Method and assumptions. For $0<s<1$, the convolution densities for one, two, and three severities are $1$, $s$, and $s^2/2$, respectively. The second follows from $\int_0^s1\,dy=s$. For the third, convolve the second density with the uniform density:

f_X^{*3}(s)=\int_0^s (s-y)\,dy=\frac{s^2}{2}, \qquad 0<s<1.

The mixture density on this interval is therefore

f_S(s)=0.2(1)+0.3s+0.4\left(\frac{s^2}{2}\right) =0.2+0.3s+0.2s^2, \qquad 0<s<1.

Now integrate this density and add the atom at zero:

P(S<1)=0.1+\int_0^1(0.2+0.3s+0.2s^2)\,ds =0.1+0.2+0.15+\frac{0.2}{3} =\frac{31}{60}=0.5167.

The same result can be checked by conditioning on the count: the contributions are $0.1$ for zero claims, $0.2(1)=0.2$ for one claim, $0.3(1/2)=0.15$ for two claims, and $0.4(1/6)=0.0667$ for three claims. Their sum is $0.5167$. Common mistake avoided: using this density formula beyond $s=1$ would be wrong; the convolution expressions change when the sum can exceed the upper limit of one individual severity.

Reading and Checking the Result

A density describes probabilities through areas, not through its value at a point. In particular, the atom $P(S=0)=p_0$ is not the value $f_S(0)$. A density may be defined differently at a single point without changing any probability obtained by integration. Keep the mass and density notation conceptually separate even when a density formula has a finite right-hand limit at zero.

For a positive threshold $s$, the CDF includes the zero-loss atom as well as the integrated positive density. Equivalently, conditioning directly on claim count gives

P(S\leq s)=p_0+\sum_{n\geq1}p_nP(X_1+\cdots+X_n\leq s\mid N=n).

This form is often convenient when only a few counts are possible or when the conditional sum probabilities are known. The density form is useful when the conditional sums have manageable convolution densities. Both are applications of the same conditioning principle.

Key takeaway: Under independent, strictly positive continuous severities, the aggregate distribution consists of an atom $P(S=0)=P(N=0)$ and a positive-loss density formed by weighting each $n$-severity convolution by $P(N=n)$. Check that the density integrates to $1-P(N=0)$.

Check Your Understanding

For each problem, distinguish the zero-loss probability from the continuous part of the aggregate distribution.

A hypothetical portfolio has $N\sim\operatorname{Bernoulli}(0.3)$ and severity uniform on $(0,4)$. Find $P(S=0)$, the density for $s>0$, and $P(S\leq2)$. (Answer: atom $0.7000$; density $0.0750$ on $(0,4)$; $P(S\leq2)=0.8500$.)
A count has distribution $\operatorname{Binomial}(2,0.4)$, and severities are uniform on $(0,1)$. Give the atom and the density on $(0,1)$ and $(1,2)$, then find $P(S\leq1)$. (Answer: atom $0.3600$; density $0.4800+0.1600s$ on $(0,1)$, and $0.1600(2-s)$ on $(1,2)$; $P(S\leq1)=0.9200$.)
A hypothetical count has probabilities $P(N=0)=0.2$, $P(N=1)=0.5$, and $P(N=2)=0.3$, with severities uniform on $(0,2)$. Find the density on each positive interval and $P(S>2)$. (Answer: atom $0.2000$; density $0.2500+0.0750s$ on $(0,2)$ and $0.0750(4-s)$ on $(2,4)$; $P(S>2)=0.1500$.)
A count takes values $0,1,2,3$ with probabilities $0.1,0.2,0.3,0.4$, respectively. Severities are uniform on $(0,1)$. Find the density for $0<s<1$ and $P(S<1)$, showing how each possible count contributes. (Answer: density $0.2000+0.3000s+0.2000s^2$; $P(S<1)=31/60=0.5167$.)

What You'll Learn